If a and n are integers and n > 1, is n odd?

If you factor, as in the post above, you can avoid cases if you arrange things differently. Rewriting Statement 1, noticing that a cannot be 1, we can factor out the smaller power, so can factor out a^(n-1) :

0 > a^n - a^(n-1)
0 > a^(n-1) (a - 1)

Now one term on the right side must be positive, and the other must be negative. But if a is a positive integer, that can't possibly be true -- both terms will be positive. So a must be negative, and (a-1) must be the negative term, and a^(n-1) must be the positive term. Since a is negative, that means the exponent must be even, so n-1 is even and n is odd.

Similarly for Statement 2, notice a cannot be -1, 0 or 1, and we have

0 > a^(3n) - a^n
0 > a^n [a^(2n) - 1]

and again, one term must be positive, the other negative. But a^(2n) has an even exponent, so it must be positive and so a^(2n) - 1 is the positive term (using the fact that a cannot be 0, 1 or -1). That makes a^n the negative term. But if a^n is negative, a must be negative, and n must be odd.

I don't think you need to do the problem algebraically, however. You can just imagine how different numbers would 'behave' in the inequalities given. So if you know that n is a positive integer greater than 1, and you know

a^(n-1) > a^n

then you might be able to see that a must be negative, since this inequality won't ever be true if a > 1. And if we know the base is negative, and we can see that one exponent is even and the other is odd (since the exponents are just consecutive integers), we know that one side of the inequality will become positive, and the other will remain negative. Clearly the larger side needs to be positive, which makes n-1 even, and n odd.

Similarly for the other Statement, the inequality

a^n > a^(3n)

simply cannot be true if a > 1. Since 1, 0 or -1 don't work in this inequality, a must be -2 or smaller. But then if n is even, it wouldn't matter if a is negative or positive, since even exponents just 'erase' our negative signs. And we know it must matter, since the inequality won't work if we have positive bases on either side. So the inequality will only work if n is odd so that the left and right sides both remain negative.

Consider substituting numbers:

statement 1 : To get stmt 1 true n=2 (even) and a = -2 (+ve numbers will never satisfy condition)

-2^1>-2^2 = -2>4 (No)

n=3 (odd) and a = -2

-2^2>-2^3 = 4>-8 (Yes)

Sufficient

Statement 2 : To get stmt 2 true n=2 (even) and a = -2 (+ve numbers will never satisfy condition)

-2^2>-2^6 = 4>64 (No)

n=3 (odd) and a = -2

-2^3>-2^9 = -8>-512 (Yes)

Sufficient

Ans D

S1) a^(n–1) > a^n
This can happen in two cases. If a is negative (e.g. -2^2 > -2^3) of if a is fraction (e.g. (1/2)^2 > (1/2)^3.
Since we are told that both a and n are integers, later case is not possible.
So we can conclude that a is negative.
Now to hold the inequality a^(n–1) > a^n true when a being negative, n-1 must be even and hence n must be odd. Sufficient

S2) In the same way as S1 this choice is also sufficient

Ans = D

Regards,

Narenn

Can someone please explain me step-wise how the factorization is done?

Thanks.

We are factoring our \(a^{n-1}\) from \(a^{n-1}-a^n\) to get \(a^{n-1}(1-a)\):

\(a^{n-1}(1-a)=a^{n-1}*1-a^{n-1}*a=a^{n-1}-a^{n-1+1}=a^{n-1}-a^{n}\).

Hope it's clear.

If we start at the "end" I can see how it is the same, however..

\(=a^{n-1}-a^{n+1}\) ?

I don't see how these two share the ^(n-1)..

We have \(a^{n-1}\) and \(a^{n}\). Now, \(a^{n}=a^{n-1}*a\).

P.S. We can factor out \(a^{n-1}\) from \(a^{n-1}-a^{n+1}\) too: since \(a^{n+1}=a^{n-1}a^2\), then \(a^{n-1}-a^{n+1}=a^{n-1}(1-a^2)\).

For more on exponents check here: math-number-theory-88376.html

Hope it helps.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.


If a and n are integers and n > 1, is n odd?

(1) a^(n–1) > a^n
(2) a^n > a^(3n)

In the original condition, there are 2 variables(a,n) and 1 equation(n>1, when it comes to inequality questions, an inequality also can be an equation.), which should match with the number equation. So you need 1 more equation. For 1) 1 equation, for 2 1equation, which is likely to make D the answer.
In 1), from a^(n-1)>a^n, if a>1, it is impossible because n-1>n impossible as well as a=0,1. So, it has to be a<0. In this case, n=odd and n-1=even. Therefore, a^(n-1)>0, a^n<0 is yes, which is sufficient.
In 2), from a^n>a^(3n), if a>1, it is impossible as well because n>3n is impossible so as a=0,1,-1. So, it has to be a<-1. In order to satisfy a^n>a^(3n) from 3n>n, only can n=odd be possible, which is yes and sufficient. Therefore, the answer is D.


-> For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.

This can be done with algebra if you're adept at factoring inequalities.

Statement 1)

\(a^{n-1}>a^n\)
\(a^{n-1}-a^n>0\)
\(a^{n-1}(1-a)>0\)
\(a^{n-1}(-1)(1-a)>0(-1) \rightarrow a^{n-1}(a-1)<0\)

If \(n\) is even then \(a^{n-1}\) is odd, and the solution to the inequality would be \(0<a<1\).
Since it's given that \(a\) is an integer, \(0<a<1\) can't be the solution, and \(n\) must be odd.

Sufficient

Statement 2)

\(a^n>a^{3n}\)
\(a^n-a^{3n}>0\)
\(a^n(1-a^{2n})>0\)
\(a^n(-1)(1-a^{2n})>0(-1) \rightarrow a^n(a^{2n}-1)<0\)
\(a^n(a^n+1)(a^n-1)<0\)

The solution to this inequality is \(a^n<-1\) and \(0<a^n<1\)
Again, it's given that \(a\) is an integer, so the second case isn't possible.
Therefore \(a^n<-1\), and \(n\) must be odd

Sufficient

Answer D

OE

(1) SUFFICIENT: We’re told that a and n are both integers and that n is greater than 1. Note further that, of n and n – 1, one of these values must be odd and one must be even. This statement indicates that, when a is raised to a smaller power (n – 1), the number becomes larger. What kind of numbers would make this true? If a is a positive integer (for example, 3), and n = 2 (recall that n must be greater than 1), then we would have 32 > 33, which is false. It’s not possible, then, to choose a = 3. Further, it’s not possible to choose any positive value for a because, if so, an–1 cannot be greater than an. (Try some additional numbers if you’re not sure.)

The value for a, then, must be negative, which triggers an interesting exception to the “typical” case where raising something to a larger power results in a larger number. When raising a negative number to a power, two different things will happen: if the power is odd, the resulting number will be negative and if the power is even, the resulting number will be positive. (If you’re not sure, test this out with some numbers: (-3)3 will stay negative, but (-3)2 will become positive.) Recall that, of n and n – 1, one must be odd and one must be even. If an–1 > an, then one of those values will be positive and one will be negative. The one on the left side of the inequality, an–1, is the larger number so it must be the positive value; therefore, n – 1 must be even. If n – 1 is even, then n itself must be odd. (Again, try some specific numbers if you’re not sure.)

(2) SUFFICIENT: We’re told that a and n are both integers and that n is greater than 1. Also note that, if n is odd, then 3n is also odd; if n is even, then 3n is also even. This statement indicates that, when a is raised to a smaller power (n), the number becomes larger. What kind of numbers would make this true? If a is a positive integer (for example, 3), and n = 2 (recall that n must be greater than 1), then we would have 32 > 36, which is false. It’s not possible, then, to choose a = 3. Further, it’s not possible to choose any positive value for a because, when a and n are positive integers, an cannot be greater than a3n.

The value for a, then, must be negative, which triggers an interesting exception to the typical case where raising something to a larger power results in a larger number. When raising a negative number to a power, two different things will happen: if the power is odd, the resulting number will be negative and if the power is even, the resulting number will be positive. (If you’re not sure, test this out with some numbers: (-3)3 will stay negative, but (-3)2 will become positive.) Test an even value for n: is (-3)2 > (-3)6 true? No; the right-hand side is larger. Further, the right-hand side will always be larger when n is even, because the two powers will always be even and the power for the right-hand number will always be larger. It is not possible, then, to choose an even value for n.

Because we know that n is an integer, we can stop here (if n cannot be even, it must be odd), but let’s also discuss the case where n is odd. Is (-3)3 > (-3)9 true? Yes! In this case, both of the resulting numbers will be negative. Because the right-hand side will always be farther away from zero, the left-hand side will always be larger. Once again, n must be odd.

The correct answer is D.

a, n are integers. Very important.


a^n - a^(n-1) < 0
=> a^(n-1) (a-1) < 0

There are two cases:
(i) a^(n-1) <0 ; a> 1
(ii) a^(n-1) > 0; a<1

No power of a positive number can make it negative. Therefore, (i) is not possible.
Only option is: a<1 and a^(n-1) > 0
Since, a is an integer, a = 0, -1, -2,.....
However, on a = 0; a^(n-1) = a^n =0, which would not fit into the inequality,
Therefore, a < 0;
But a^(n-1) > 0
Thus, n-1 is even
Or, n is odd

Statement 1 is sufficient



a^3n - a^n < 0
a^n ((a^2n) - 1) < 0;

Two possibilities:

(i) a^n<0; a^2n>1
(ii) a^n>0; a^2n - 1 <0

a /= 0, since the inequality won't stand if a is zero
and for any non zero integral value of a, (a^2n - 1) can never be negative
Thus, Case (ii) is out
Therefore, a^n < 0
Only possible if a<0 and n is odd
Thus, Statement 2 is sufficient

Therefore, (D)

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